# Annuity that pays $t^2$ at time $t$ in arrears annually.

I am asked to show that such an annuity for $n$ years will be expressed as,

$$\frac{2(Ia)_{\bar n|} - a_{\bar n|}-n^2u^{n+1}}{1-u}$$

where $u=\frac{1}{1+i}$ and $i$ is the annual effective interest rate.

So essentially the sum

$u+4u^2+9u^3+...+nu^n$. My idea was $u+(2u^2+2u^2)+(3u^3+3u^3+3u^3)+...$ and make a series of annuities which we are familiar with e.g. Increasing annuity, immediate annuity, but this only made it super complicated. But I can't think of any other way that might lead to solving this.

Any help? Does anyone know how to prove this?

Thanks a lot

• Do you know/are you allowed to use calculus? Because there's a slick solution that involves taking derivatives, and a somewhat messier solution along the lines you've laid out... – Micah Dec 11 '15 at 18:17
• Hi there, I think there are no restrictions to the method itself as long as it's got logic into it and gets us to the right answer! calculus was not in my mind and sounds interesting(not sure if I'll fully understand though), can you show me please ? thank you! – John Trail Dec 11 '15 at 20:04

Let us denote such an annuity as $\bar S$.

Then $$\bar S = u + 4u^2 + 9u^3 +\cdots +n^2u^n \tag{1}$$ Now multiply (1) by u ,

$$u\bar S = u^2 + 4u^3 + 9u^4 +\cdots +n^2u^{n+1} \tag{2}$$

Substract (1)-(2),

$$(1-u)\bar S = u+3u^2+5u^3+7u^4+\cdots+(2n-1)u^n - n^2u^{n+1}$$

$$(1-u)\bar S = (2u-u)+(4u^2-u^2)+(6u^3-u^3)+\cdots+(2nu^n-u^n) - n^2u^{n+1}$$

$$(1-u)\bar S = 2(u+2u^2+3u^3+\cdots+nu^n)-(u+u^2+u^3+\cdots+u^n)-n^2u^{n+1}$$

The terms in the first bracket constitute an increasing annuuity with P = 1 and D = 1 and the terms in the next bracket constitute an immediate annuity. Thus if you rearrange terms, you get

$$(1-u)\bar S = 2(Ia)_{\bar n|} - a_{\bar n|}-n^2u^{n+1}$$

$$\bar S = \frac{2(Ia)_{\bar n|} - a_{\bar n|}-n^2u^{n+1}}{1-u}$$